Problem: Solve for $x$ : $ 3|x - 3| + 1 = -2|x - 3| + 3 $
Answer: Add $ {2|x - 3|} $ to both sides: $ \begin{eqnarray} 3|x - 3| + 1 &=& -2|x - 3| + 3 \\ \\ { + 2|x - 3|} && { + 2|x - 3|} \\ \\ 5|x - 3| + 1 &=& 3 \end{eqnarray} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} 5|x - 3| + 1 &=& 3 \\ \\ { - 1} &=& { - 1} \\ \\ 5|x - 3| &=& 2 \end{eqnarray} $ Divide both sides by ${5}$ $ \dfrac{5|x - 3|} {{5}} = \dfrac{2} {{5}} $ Simplify: $ |x - 3| = \dfrac{2}{5}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{2}{5} $ or $ x - 3 = \dfrac{2}{5} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{2}{5} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{2}{5} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{2}{5} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $5$ $ x = - \dfrac{2}{5} {+ \dfrac{15}{5}} $ $ x = \dfrac{13}{5} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{2}{5} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{2}{5} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{2}{5} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $5$ $ x = \dfrac{2}{5} {+ \dfrac{15}{5}} $ $ x = \dfrac{17}{5} $ Thus, the correct answer is $x = \dfrac{13}{5} $ or $x = \dfrac{17}{5} $.